Question

In a system using the fixed partitions memory allocation scheme, given the following situation (and using decimal form): After Job J is loaded into a partition of size 50K, the resulting fragmentation is 7168 bytes. What is the size of Job J in bytes

Answer 1

Answer:

Answer is 51200 bytes.

Therefore, refer below.

Explanation:

The size of Job J in bytes is 51200.

Answer 2

Following are the calculation to the size of Job J in bytes:

Assuming that the fixed division is supplied 50 K:

$\to 50 \times 1024\ Bytes =51200\ bytes\\\\ \therefore \\\\ (1\ K=1024 \ bytes)$

After allocating the partitions to the work, the unused byte (Fragmentation) is .

J's size is calculated as follows:

$\to 51200-7168=44032 \ bytes$

Therefore, the answer is "$44032 \ bytes$".

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